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<math>x(t)= \sum^{\infty}_{k = -\infty} x(kT) (u[t-kT]-u[t-(k+1)T])</math> | <math>x(t)= \sum^{\infty}_{k = -\infty} x(kT) (u[t-kT]-u[t-(k+1)T])</math> | ||
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2. First-order intapolation | 2. First-order intapolation | ||
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where <math>f_k (t)= x(t_k) + (t-t_k) \frac {x(t_{k+1})-x(t_k)}{t_{k+1} - t_k} for t_k < t < t_{k+1} </math> | where <math>f_k (t)= x(t_k) + (t-t_k) \frac {x(t_{k+1})-x(t_k)}{t_{k+1} - t_k} for t_k < t < t_{k+1} </math> | ||
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Latest revision as of 09:47, 10 November 2008
Methods to recover a signal
1. Zero-order intapolation (step function)
$ x(t)= \sum^{\infty}_{k = -\infty} x(kT) (u[t-kT]-u[t-(k+1)T]) $
2. First-order intapolation
$ x(t)= \sum^{\infty}_{k = -\infty} f_k (t) $
where $ f_k (t)= x(t_k) + (t-t_k) \frac {x(t_{k+1})-x(t_k)}{t_{k+1} - t_k} for t_k < t < t_{k+1} $
